Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 1698

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

Short Answer

Expert verified
The maximum possible power consumed by the resistor is \(2 \mathrm{~W}\), which corresponds to option (D).

Step by step solution

01

Calculate the total EMF and internal resistance of the circuit

The batteries are connected in series, so the total EMF of the circuit is the sum of the emf of both batteries, and the total internal resistance is the sum of their internal resistances. Total EMF (E_total) = E_1 + E_2 = 2V + 2V = 4V Total internal resistance (r_total) = r_1 + r_2 = 1Ω + 1Ω = 2Ω
02

Apply the maximum power theorem

To maximize the power consumed by the resistor, we need to find the value of R when the power is at its maximum. According to the maximum power transfer theorem, this occurs when the load resistance (R) is equal to the internal resistance (r_total) of the circuit. So, R = r_total = 2Ω
03

Calculate the total current in the circuit

Now, let's find the total current in the circuit (I_total) using Ohm's law: I_total = E_total / (R + r_total) = 4V / (2Ω + 2Ω) = 4V / 4Ω = 1A
04

Calculate the maximum power consumed by the resistor

Finally, let's calculate the maximum power (P_max) consumed by the resistor using the formula P = I^2 * R: P_max = I_total^2 * R = (1A)^2 * 2Ω = 1A * 2Ω = 2W So the maximum possible power consumed by the resistor is 2 W, which corresponds to option (D).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Assertion and reason are given in following questions each question has four options one of them is correct select it. (a) Both assertion and reason are true and the reason is correct reclamation of the assertion. (b) Both assertion and reason are true, but reason is not correct explanation of the assertion. (c) Assertion is true, but the reason is false. (d) Both, assertion and reason are false. Assertion: There is no current in the metals in the absence of electric field. Reason: Motion of free electrons is random. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

A bulb of \(300 \mathrm{~W}\) and \(220 \mathrm{~V}\) is connected with a source of \(110 \mathrm{~V}\). What is the \(\%\) decrease in power? (A) \(100 . \%\) (B) \(75 \%\) (C) \(70 \%\) (D) \(25 \%\)

Area of cross-section of two wires of same length carrying same current is in the ratio of \(1: 2\). Then the ratio of heat generated per second in the wires \(=\ldots\) (A) \(1: \sqrt{2}\) (B) \(1: 1\) (C) \(1: 4\) (D) \(2: 1\)
See all solutions

Recommended explanations on English Textbooks

Key Concepts in Language and Linguistics

Read Explanation

Lexis and Semantics

Read Explanation

Argumentative Essay

Read Explanation

Semiotics

Read Explanation

Global English

Read Explanation

Single Paragraph Essay

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free